Question

A block of mass N is attached with a spring of spring constant K whose other end is aatached with a wall. Initially the spring is in natural length and all the surfaces are friction less then the maximum compression in the spring if a bullet of mass m and velocity U gets embedded in the block

Answer 1

Hello ,

by conservation of momentum;

mU+N*0= (m+N)*X --- X is new velocity of the spring + block + bullet system after collision

which means

mU=(m+N)X ;

also :by conservation of kinetic energy

mU^2 = (m+N)X^2

: X^2 = [mU^2]/(m+N) --------eq(1)

Now applying work enégy theorm

½ k(x)^2 = ½(m+N)X^2

>>kx^2 = ( m+N) * mU^2/(m+N)

= mU^2

>>>x^2 = mU^2/k

therefore::

x = U* [m/k]^½